???? 前言
大家好呀，我是毛小悠，能够叫我二毛，在家中排行老二，是一名前端开发工程师。
本系列文章旨在通过练习来进步JavaScript的能力，一起欢快的做题吧。????????????
以下每道题，二毛我都有尝试做一遍。倡议限时训练，比方限定为半小时，如果半小时内想不进去，能够联合文章开端的参考答案来思考。
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???? 题目1：二进制搜寻树验证
一个二叉搜寻树 是有序的二叉树。这意味着，如果要应用有序遍历将树转换为数组，则该数组将按排序程序。这种排序的益处是，当树均衡时，搜寻是对数工夫操作，因为您查看的每个节点都不是您要搜寻的节点，因而您能够抛弃搜寻树的另一半。
您将编写一个函数来验证给定的二叉树是否为二叉搜寻树。排序程序不是预约义的，因而两者均可应用。
这些是无效的二进制搜寻树：
5 / \ 2 7 / \ \1 3 9 7 / \9 2这些不是：
1 / \2 3 5 / \2 9 \ 7习题代码：
// This is here as documentation. The nodes in the tree are instances of// this class. You don't need to change this implementation.class Node { constructor(value, left = null, right = null){ this.value = value; this.left = left; this.right = right; }}const isBST = node => { // Your solution here. return false;};???? 题目2：简略的乐趣：原始编号
工作
约翰有一个重要的数字，他不心愿其他人看到它。
他决定应用以下步骤对数字进行加密：
His number is always a non strict increasing sequenceie. "123"He converted each digit into English words.ie. "123"--> "ONETWOTHREE"And then, rearrange the letters randomly.ie. "ONETWOTHREE" --> "TTONWOHREEE"约翰感觉这样做很平安。实际上，这种加密能够很容易地解密:(
给定加密的字符串s，您的工作是解密它，并以字符串格局返回原始数字。
留神，您能够假设输出字符串s始终无效；它仅蕴含大写字母；解密后的数字按升序排列；前导零是容许的。
例
对于s =“ ONE”，输入应为1。
对于s =“ EON”，输入也应为1。
对于s =“ ONETWO”，输入应为12。
对于s =“ OONETW”，输入也应该为12。
对于s =“ ONETWOTHREE”，输入应为123。
对于s =“ TTONWOHREEE”，输入也应该为123。
习题代码：
function originalNumber(s){ //coding and coding.. }答案
???? 题目1的答案
参考答案1：
class Node { constructor(value, left = null, right = null) { this.value = value; this.left = left; this.right = right; }}function isBST(node) { const g = inOrder(node); const a = g.next(); if (a.done) return true; const b = g.next(); if (b.done) return true; var p = b.value; const asc = a.value < p; for (const v of g) { if (asc ? p > v : p < v) return false; p = v; } return true;}function* inOrder(n) { if (n !== null) { yield* inOrder(n.left); yield n.value; yield* inOrder(n.right); }}参考答案2：
class Node { constructor(value, left = null, right = null) { this.value = value; this.left = left; this.right = right; }}const isMinFirstBST = (node, min, max) => { return ( (min === undefined || min < node.value) && (max === undefined || node.value < max) && (node.left === null || isMinFirstBST(node.left, min, node.value)) && (node.right === null || isMinFirstBST(node.right, node.value, max)));};const isMaxFirstBST = (node, min, max) => { return ( (min === undefined || node.value > min) && (max === undefined || max > node.value) && (node.left === null || isMaxFirstBST(node.left, node.value, max)) && (node.right === null || isMaxFirstBST(node.right, min, node.value)));};const isBST = (node, min, max) => { return node === null || isMinFirstBST(node) || isMaxFirstBST(node);};参考答案3：
// This is here as documentation. The nodes in the tree are instances of// this class. You don't need to change this implementation.class Node { constructor(value, left = null, right = null){ this.value = value; this.left = left; this.right = right; }}const isBST = node => { const list = inOrder(node); return isSorted(list, 1) || isSorted(list, -1);};const inOrder = node => { if (node == null) { return []; } if (node instanceof Node) { return [...inOrder(node.left), node.value, ...inOrder(node.right)]; } return [node];}const isSorted = (list, order) => { return list.every((x,i) => i == 0 || Math.sign(x - list[i - 1]) == order);}参考答案4：
// This is here as documentation. The nodes in the tree are instances of// this class. You don't need to change this implementation.class Node { constructor(value, left = null, right = null){ this.value = value; this.left = left; this.right = right; }}const isBST = node => { const arr = inOrder(node); return arr.every( (v, i, a) => i == 0 ? true : v > a[i-1]) || arr.every( (v, i, a) => i == 0 ? true : v < a[i-1]); function inOrder(node) { if (node == undefined) return []; return inOrder(node.left).concat(node.value).concat(inOrder(node.right)); }};参考答案5：
function Node(v,l=null,r=null) { this.value = v, this.left = l, this.right = r; }const ino = (n,a=[]) => (n.left && ino(n.left,a), a.push(n.value), n.right && ino(n.right,a), a), isBST = n => !n || ino(n).every((v,i,a) => !i || (a[0]<a[1] ? a[i-1]<v : v<a[i-1]));???? 题目2的答案
参考答案1：
function originalNumber(s){ let digits = [] s = s.split(''); [ ['Z','ZERO',0], ['W','TWO',2], // Order matters! ['G','EIGHT',8], ['X','SIX',6], ['S','SEVEN',7], // For example, must count and splice all SEVENs ['V','FIVE',5], // _before_ using 'V' to search for FIVEs ['F','FOUR',4], ['H','THREE',3], ['O','ONE',1], ['I','NINE',9] ].forEach(([unique,all,d]) => { let n = s.filter(c => c == unique).length for (let i = 0; i < n; i++) { all.split('').forEach(c => s.splice(s.indexOf(c),1)) digits.push(d) } }) return digits.sort().join``}参考答案2：
function originalNumber(s) { const r = [] const d = {E: 0, F: 0, G: 0, H: 0, I: 0, N: 0, O: 0, R: 0, S: 0, T: 0, U: 0, V: 0, W: 0, X: 0, Z: 0} for (const c of s) ++d[c] for (; d.Z; d.Z--, d.O--) r.push(0) for (; d.W; d.W--, d.O--) r.push(2) for (; d.U; d.F--, d.O--, d.U--) r.push(4) for (; d.F; d.F--, d.I--, d.V--) r.push(5) for (; d.X; d.I--, d.X--) r.push(6) for (; d.V; d.V--) r.push(7) for (; d.G; d.I--, d.G--, d.H--) r.push(8) for (; d.I; d.I--) r.push(9) for (; d.O; d.O--) r.push(1) for (; d.H; d.H--) r.push(3) return r.sort().join("")}参考答案3：
function originalNumber(s){ const c=l=> ( s.match(new RegExp(l,'g')) || []).length; return '0'.repeat(c('Z')) + '1'.repeat(c('O')- c('Z')- c('W') - c('U') ) + '2'.repeat(c('W')) + '3'.repeat(c('H')-c('G')) + '4'.repeat(c('U')) + '5'.repeat(c('F')-c('U')) + '6'.repeat(c('X')) + '7'.repeat(c('V')-c('F')+ c('U')) + '8'.repeat(c('G')) + '9'.repeat(c('I')-c('F')+ c('U')-c('X')-c('G')) ;}参考答案4：
const originalNumber = (_ => { let search = [ [ 0, 'Z', 'ZERO' ], [ 8, 'G', 'EIGHT' ], [ 6, 'X', 'SIX' ], [ 4, 'U', 'FOUR' ], [ 5, 'F', 'FIVE' ], [ 7, 'V', 'SEVEN' ], [ 9, 'I', 'NINE' ], [ 2, 'W', 'TWO' ], [ 3, 'H', 'THREE' ], [ 1, 'N', 'ONE' ] ]; return str => { let res = [] , freq = [ ...str ].reduce((a, b) => (a[b] = (a[b] || 0) + 1, a), {}) , count = 0; for (let [ value, key, word ] of search) { let count = freq[key]; if (!count) continue; for (let char of word) freq[char] -= count; while (count--) res.push(value); } return res.sort((a, b) => a - b).join(''); }})();参考答案5：
function originalNumber(s) { const pairs = { 0: {string: 'ZERO', key: 'Z', num: 0}, 1: {string: 'TWO', key: 'W', num: 2}, 2: {string: 'FOUR', key: 'U', num: 4}, 3: {string: 'SIX', key: 'X', num: 6}, 4: {string: 'EIGHT', key: 'G', num: 8}, 5: {string: 'ONE', key: 'O', num: 1}, 6: {string: 'THREE', key: 'H', num: 3}, 7: {string: 'FIVE', key: 'F', num: 5}, 8: {string: "SEVEN", key: 'S', num: 7}, 9: {string: 'NINE', key: 'N', num: 9} }; let decoded = []; Object.keys(pairs).forEach(function (key) { // for each key in pairs object if (s.includes(pairs[key].key)) { // includes returns true/false, based on if char is in string let result = contains(s, pairs[key], decoded); s = result[0]; decoded = result[1]; } }); return decoded.sort().join('').toString();}function contains(s, pair_index, decoded) { // removes char at index while (s.includes(pair_index.key)) { // if string contains object key, remove full number decoded.push(pair_index.num); for (let i = 0; i < pair_index.string.length; i++) { s1 = s.slice(0, s.indexOf(pair_index.string[i])); s2 = s.slice(s.indexOf(pair_index.string[i]) + 1); s = s1 + s2; } } return [s, decoded];}????后序
本系列会定期更新的，题目会由浅到深的逐步提高。
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